Question

Find the coordinates of the foci, the vertices the eccentricity and the length of the latus rectum of the hyperbola $5y^2-9x^2=36$

Answer 1

The given equation is $5y^2-9x^2=36$
$\Rightarrow$ $\frac{y^2}{\left(\frac{36}{5}\right)}-\frac{x^2}{4}=1$
$\Rightarrow \frac{y^2}{{\left(\frac{6}{\sqrt{5}}\right)}^2}-\frac{x^2}{2^2}=1...\left(1\right)$
On comparing equation (1) with the standard equation of hyperbola
i.e., $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$,
we obtain $a=\frac{6}{\sqrt{5}}$ and $b=2$
We know that $a^2+b^2=c^2,$ where $c=ae$
$\therefore c^2=\frac{36}{5}+4=\frac{56}{5}$
$\Rightarrow c=\sqrt{\frac{56}{5}}=\frac{2\sqrt{14}}{\sqrt{5}}$
Therefore, the coordinates of the foci are $(0,\pm \frac{2\sqrt{14}}{\sqrt{5}})$
The coordinates of the vertices are $(0,\pm \frac{6}{\sqrt{5}})$
Eccentricity $e=$ $\frac{c}{a}=\frac{\left(\frac{2\sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}$
Length of latus rectum $=$ $\frac{2b^2}{a}=\frac{2\times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4\sqrt{5}}{3}$